what’s wrong here

[mighty flower]

???

[keen hound]

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[weary summit]

take 256 common

add 1/2 ^ 3 to 8

whats the confusion?

maybe u complicated it unneseccarily

take a = 1/2 and r = 1/2

[mighty flower]

Well I was looking for the sum but got 42 which isn’t right

[weary summit]

u know formula for sum of geometric sequence?

[mighty flower]

I think so that’s what I tried to do

[weary summit]

so get sum till 8 terms and subtract sum of 1/2 and 1/4

[mighty flower]

I plugged in 64 for a which is wrong but heres my new work which is still wrong

[weary summit]

caus it's from 3rd to 8th term

[weary summit]

why did u plug 64 , take 256 common

work with 1/2

[mighty flower]

But it starts at term theee

Which is 32

So I was counting 32 as the first term and the eighth term as the sixth

[weary summit]

yes but what im saying is u can

1. find sum of series where 1/2 is a and 1/2 is r till 8 terms
2. subtract 1/2+1/4
   3)multiply the whole thing by 256

[mighty flower]

Oh ok

But what did I do wrong in my work

[weary summit]

it's just so lengthy it's bound to calculation errors

i cant be bothered to find it

but there is some mistake fs

[mighty flower]

Alr

[timid talon]

let me see

$\sum_{j=3}^{8} 256(\frac{1}{2})^j$

[steep tulip]

;(

[timid talon]

so geom series formula is

$\frac{a(1-r^{n+1})}{1-r}$

a=32

r=1/2

[steep tulip]

;(

[weary summit]

but we can take out 256 common

[timid talon]

bro let me cook

[weary summit]

im just giving u an ingrediant

[timid talon]

$\frac{32(1-(\frac{1}{2})^6)}{1-\frac{1}{2}}$

[steep tulip]

;(

[timid talon]

so $64(1-\frac{1}{64})$

[steep tulip]

;(

[timid talon]

and its 63

[timid talon]

its not -1/2, its just 1/2

let me check wolfram

,w sum_{j=3}^{8} 256(\frac{1}{2})^j

[steep tulip]

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[timid talon]

yep

its correct

@mighty flower look through my work

[mighty flower]

Thank you so much

+close